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3 years ago

15

Solar cells are often coated with a transparent, thin film of silicon monoxide (n = 1.45) to minimize reflective losses from the

surface. Suppose a silicon solar cell (n = 3.5) is coated with a thin film of silicon monoxide. Determine the minimum film thickness that produces the least reflection at a wavelength of 550 nm.

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

94.82 nm

Explanation:

We have given that wavelength $\lambda=550\ nm$

We have to find the minimum film thickness that produces the least reflection

minimum thickness is given by $t=\frac{\lambda }{4n}$

here n is given n=1.45

so minimum thickness $t=\frac{550 }{4\times 1.45}=94.82\ nm$

so the minimum film thickness will be 94.82 nm

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Which law is used to find the magnitude of a magnetic force?

Talja [164]

Answer:

The Flemings left hand rule is used to find the magnitude of a magnetic force

Explanation:

Fleming's left hand rule states that if the first three fingers are held mutually at right angles to one another, then the fore finger points into the direction of magnetic field the middle finger in the direction of current while the thumb points in the direction of force.

Mathematically

Magnetic Force F= BILsinθ

Where

B= magnetic field density Tesla

I= current

L= length of conductor

θ= angle of conductor with field

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3 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

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3 years ago

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

So

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

So

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

So

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

So

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.

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3 years ago

which drawing below represents the electric field surrounding two objects that have equal magnitude changes of opposite polarity

Fantom [35]

Answer:

the 3rd one

Explanation:

j cause I think so

8 0

2 years ago

Two fishing boats depart a harbor at the same time, one traveling east, the other south. the eastbound boat travels at a speed 1

Novosadov [1.4K]

<span>they are travelling at right angles to each other.
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</span> East bound <span> = x+1 [mi/h]
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d=5x
and
d=5(x+1)
miles away from the starting point
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25x^2+(5x+5)^2=625
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50</span>x^2+50x-600=0
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(x+4)(x-3)=0
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3 years ago